Asolution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 325 ml of solution. how many moles of ammonium chloride are present in the resulting solution

0.206 moles if question is actually correctly worded (unlikely) 0.633 M if question should have asked about molarity (likely). First, let's calculate how many moles of ammonium chloride (NH4Cl) we have. Start by looking up the atomic weights of involved elements. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight chlorine = 35.453 Molar mass NH4Cl = 14.0067 + 4*1.00794 + 35.453 = 53.49146 g/mol Moles NH4Cl = 11.0 g / 53.49146 g/mol = 0.205640302 moles Now technically, I already have the answer to this problem in that there is 0.205640302 moles of ammonium chloride in the solution. Doesn't matter if 325 mL were made, or 1000 liters. The total number of moles of ammonium chloride remains the same. But I suspect that this question was improperly worded and that what is actually being looked for is the molarity of the resulting solution. And since molarity is defined as moles per liter, a bit of division is needed. So: 0.205640302 mol / 0.325 L = 0.632739391 mol/L = 0.632739391 M Rounding results to 3 significant figures gives 0.206 moles and 0.633 M.